Let $E$ be a path-connected subset of $\mathbb{R^3}$, and $\gamma$ a closed curve in $E$ (not necessarily simple). If I know that $\gamma$ is homotopic to a point trough a closed homotopy inside $E$, is $\gamma$ the boundary of a $2$-manifold contained in $E$?
Intuition seems to be so. Surely, I'm not counting for some pathological cases that I should be (i.e., when $E$ is a line, it can't contain a 2-manifold, right?). But the image of the homotopy is a candidate for such a region, if I can only guarantee that exists an homotopy whose boundary will be exactly $\gamma$ (which seems reasonable, for it informally works on all the well-behaved regions I could think of). The point is to make such argument work:
Let $\gamma_1$ and $\gamma_2$ be two closed homotopic curves in $E$, and let $\omega$ be a closed $1$-form in $E$. Then: $$\int_{\gamma_1} \omega = \int_{\gamma_2} \omega$$ PROOF: If $\gamma_1$ and $\gamma_2$ are homotopic, then $\gamma_1 - \gamma_2$ is homotopic to a point. It is then the boundary of a surface $M$ on $E$: $\partial M = \gamma_1 - \gamma_2$. Applying the Stokes' theorem gives: $$\int_{\gamma_1 - \gamma_2} \omega = \int_{\partial M} \omega = \int_{M} d\omega = \int_M 0 = 0$$ By additivity of the integral sign, we have: $$\int_{\gamma_1 - \gamma_2} \omega = \int_{\gamma_1} \omega - \int_{\gamma_2} \omega$$ Combining both gives the result.
I found those related questions: here and here. Seems that I need to require $E$ to be open. It would exclude the line example in this case, but I would need to account for all the cases when $E$ isn't open (though I think it works if I separate it in interior and boundary points).